Higher Prelim Topic Revision - Web Version

Higher Formula Sheet

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1 Straight Line

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Hint

Perpendicular bisectors pass through the midpoint of a line segment, and run perpendicular to them.

Answer

\(4y=-3x+29\) or \(y=-\frac{3}{4}x+\frac{29}{2}\)

For more practice, try: Zeta Higher Textbook, Page 15, Exercise 1.8, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

A straight line equation must be rearranged to make \(y\) the subject (\(y=\dots\)) before its gradient can be determined.

Answer

\(2y=-x-5\) or \(y=-\frac{1}{2}x-\frac{5}{2}\)

For more practice, try: Zeta Higher Textbook, Page 13, Exercise 1.6, Questions 3, 4 and 5

Solution

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Hint

The formula \(m=\tan{\theta}\) links a line's gradient to the angle it makes with the positive direction of the \(x\)-axis.

Answer

\(y=-x+2\)

For more practice, try: Zeta Higher Textbook, Page 9, Exercise 1.4A, Questions 1, 2 and 3

Solution

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Hint

Calculate the gradients of PQ and QR first. The conclusion to a collinearity problem needs to be learned carefully.

Answer

\(m_{PQ}=m_{QR}=-3\), valid statement (see solution).

For more practice, try: Zeta Higher Textbook, Page 12, Exercise 1.5, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

1. Medians meet opposite sides at the midpoint.

2. Altitudes meet opposite sides at an angle of 90 degrees.

3. You can use \(y=y\) as a first step to finding points of intersection.

Answer

1. \(y=-3x+5\)

2. \(2y=-x-5\) or \(y=-\frac{1}{2}x-\frac{5}{2}\)

3. \((3,-4)\)

For more practice, try: Zeta Higher Textbook, Page 21, Exercise 1.11B, Questions 1, 2 and 3

Solution (a)

Solution (b)

Solution (c)

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2 Recurrence Relations

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Hint

1. Calculate \(u_1\) first using the value of \(u_0\) as a starting point.

3. What is the value of \(a\)?

2. What is the formula for the limit of a recurrence relation?

Answer

1. \(15\)

3. \(-1<a<1\), valid statement (see solution).

2. \(9\)

For more practice, try: Zeta Higher Textbook, Page 90, Exercise 6.1, Questions 1(a), 1(b) and 1(c)

Solution (a)

Solution (b)

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Hint

1. Can you put both \(u_2\) and \(u_3\) into the recurrence relation? Where does each one go?

2. What is the value of \(a\)?

Answer

1. \(-4\)

2. \(a>1\), valid statement (see solution).

For more practice, try: Zeta Higher Textbook, Page 93, Exercise 6.3, Questions 1(f), 1(i) and 1(l)

Solution (a)

Solution (b)

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Hint

1. Substitute the value of \(u_4\) into the recurrence relation.

2. Use your expression from part (a), and the new information that \(u_5=-1\).

Answer

1. \(6k-4\)

2. \(k=\frac{1}{2}\)

For more practice, try: Zeta Higher Textbook, Page 99, Exercise 6.6B, Questions 9, 10 and 11

Solution (a)

Solution (b)

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Hint

1. What "12% is subtracted (from the original 100%) as a decimal?

2. Since dividing by \(0.12\) is tricky, multiply both parts of the fraction by \(100\) and work from there.

Answer

1. \(a=0.88,b-30\)

2. 250 squirrels

For more practice, try: Zeta Higher Textbook, Page 96, Exercise 6.5B, Questions 1, 2 and 3

Solution (a)

Solution (b)

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Hint

Create two equations: one linking the first two terms, and a second linking the second term and the third term. Solve simultaneously.

Answer

\(p=-\frac{1}{2},q=2\)

For more practice, try: Zeta Higher Textbook, Page 91, Exercise 6.2, Questions 1(a), 1(b) and 1(c)

Solution

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3 Differentiation I

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Hint

To differentiate, multiply by the power and then reduce the power by 1...

Answer

\(-19\)

For more practice, try: Zeta Higher Textbook, Page 155, Exercise 11.4A, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

Start by splitting the fraction into two fractions, each with a denominator of \(3x^2\), and prepare each part for differentiation.

Answer

\(2x^2+2x^{-3}\)

For more practice, try: Zeta Higher Textbook, Page 155, Exercise 11.3B, Questions 2(e), 2(i) and 2(m)

Solution

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Hint

The gradient of a tangent to a curve is described by its derivative, \(\dfrac{dy}{dx}\).

Answer

\(9\)

For more practice, try: Zeta Higher Textbook, Page 156, Exercise 11.4A, Questions 2(a), 2(c) and 2(e)

Solution

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Hint

The gradient of a tangent was found in the last question. To find its equation, we also need to know a coordinate the tangent passes through.

Answer

\(y=6x-14\)

For more practice, try: Zeta Higher Textbook, Page 158, Exercise 11.5, Questions 1(a), 2(a) and 3(a)

Solution

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Hint

The rate of change of a function is described by its derivative, \(h'(t)\). A square root \(\sqrt{x}\) can also be written in index as \(x^{\frac{1}{2}}\).

Answer

\(2\)

For more practice, try: Zeta Higher Textbook, Page 157, Exercise 11.4B, Questions 1, 2 and 7

Solution

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Hint

Find the derivative, \(\dfrac{dy}{dx}\), first.

If a value of \(x\) is substituted into the derivative, the result will be the gradient. Here, when \(x=2\) is substituted in, what value would the output be equal to?

Answer

\(k=-\frac{1}{2}\)

No similar questions found in the Zeta textbook. Try this SQA exam question from 2024.

Solution

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4 Quadratic Theory

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Hint

A few methods may be used.

One begins by fully expanding \(p(x+q)^2+r\).

Answer

\(3(x-4)^2+9\)

For more practice, try: Zeta Higher Textbook, Page 33, Exercise 2.6D, Questions 1, 2 and 3

Solution

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Hint

A few methods may be used.

One begins by fully expanding \(a(x+b)^2+c\).

Answer

\(-(x-2)^2+7\)

For more practice, try: Zeta Higher Textbook, Page 32, Exercise 2.6C, Questions 1, 2 and 3

Solution

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Hint

Start by considering the roots of the equation \(2x^2+8x-10\). Notice the common factor.

You need to draw a sketch.

Answer

\(x<-5,x>1\)

For more practice, try: Zeta Higher Textbook, Page 37, Exercise 2.10, Questions 7, 10 and 13

Solution

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Hint

Start by considering the roots of the equation \(m^2-m-20\).

You need to draw a sketch.

Since \(m\) is the variable, your answer must be in terms of \(m\).

Answer

\(-4\leqslant m\leqslant5\)

For more practice, try: Zeta Higher Textbook, Page 37, Exercise 2.10, Questions 11, 16 and 20

Solution

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Hint

An equation has (real) equal roots when \(b^2-4ac=0\).

Answer

\(p=\pm6\)

For more practice, try: Zeta Higher Textbook, Page 39, Exercise 2.12, Questions 1(a), 1(b) and 1(e)

Solution

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Hint

An equation has no real roots when \(b^2-4ac<0\).

Answer

\(q<-\frac{6}{5}\)

For more practice, try: Zeta Higher Textbook, Page 39 Exercise 2.12, Questions 2(a) and 3(a)

Solution

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Hint

An equation has real, distinct roots when \(b^2-4ac>0\).

Answer

\(k<-2,k>6\)

For more practice, try: Zeta Higher Textbook, Page 39, Exercise 2.12, Questions 2(c), 2(d) and 3(f)

Solution

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Hint

Points of intersection can be found by solving \(y=y\)...

Answer

\((-2,1)\) and \((4,5)\)

For more practice, try: Zeta Higher Textbook, Page 41, Exercise 2.15, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

Tangency can be explored by solving \(y=y\) are for generally finding points of intersection, then considering how the resulting equation can be factorised or \(b^2-4ac\).

Answer

\((2,5)\)

For more practice, try: Zeta Higher Textbook, Page 41, Exercise 2.15, Questions 2(a), 2(b) and 2(c)

Solution

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5 Sets and Functions

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Hint

Start by writing it in the form \(y=...\) and then rearrange to make \(x\) the subject.

Answer

\(h^{-1}(x)=3x+4\)

For more practice, try: Zeta Higher Textbook, Page 73, Exercise 4.4, Questions 1(a), 1(e) and 1(i)

Solution

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Hint

Start by writing it in the form \(y=...\) and then rearrange to make \(x\) the subject.

Answer

\(f^{-1}(x)=\left(\dfrac{x+3}{2}\right)^2\)

For more practice, try: Zeta Higher Textbook, Page 73, Exercise 4.4, Questions 2(b), 2(d) and 2(i)

Solution

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Hint

Division by zero is undefined.

Answer

\(x\ne \frac{1}{2}\)

For more practice, try: Zeta Higher Textbook, Page 67, Exercise 4.1A, Questions 1(b), 1(d) and 1(g)

Solution

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Hint

The square root of a negative is undefined.

Answer

\(x<3\)

For more practice, try: Zeta Higher Textbook, Page 67, Exercise 4.1A, Questions 2(b), 2(d) and 2(k)

Solution

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Hint

The expression for \(g(x)\) needs to be substituted into the expression for \(f(x)\).

Answer

\(f(g(x))=2x-1\)

For more practice, try: Zeta Higher Textbook, Page 70, Exercise 4.3A, Questions 1(a), 1(b) and 1(g)

Solution

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Hint

The expression for \(f(x)\) needs to be substituted into the expression for \(g(x)\).

Simplify it fully.

2. Applying \(f\) to \(x\) and then applying \(g\) to the result leads back to \(x\). What is happening here?

Answer

1. \(g(f(x))=x\)

2. \(f\) and \(g\) are inverse to each other.

No similar questions found in the Zeta textbook. Try this SQA exam question from 1993(!), or this one from the 2018 specimen paper.

Solution

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Hint

1. Substitute \(g(x)\) into \(f(x)\) and simplify.

2. Observe the results from part (a) before simplifying, when it when written in completed square form, \((x-a)^2+b\). What is the lowest possible value of \(k(x)\)?

3. Begin by factorising \(k(x)\).

Answer

1. \(k(x)=x^2-6x+8\)

2. \(k(x)\geqslant-1\).

For more practice, try: Zeta Higher Textbook, Page 68, Exercise 4.1B, Questions 1(f), 1(h) and 1(l)

Solution

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6 Trigonometry

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Hint

Rearrange to make \(\sin x\) the subject. Give your final answer in radians.

Answer

\(x=\frac{\pi}{6},\frac{5\pi}{6}\)

For more practice, try: Zeta Higher Textbook, Page 122, Exercise 8.5A, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

Use a CAST diagram carefully to identify quadrants in which \(\tan x\) is negative.

Answer

\(x=\frac{5\pi}{12},\frac{11\pi}{12}\)

For more practice, try: Zeta Higher Textbook, Page 122, Exercise 8.5A, Questions 1(g), 1(h) and 1(i)

Solution

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Hint

1. At which angle is \(y=4\cos x\) at its minimum, and how low does it go? Answer in radians.

2. The points of intersection Q and R can be obtained by solving the set of equations \(y=4\cos x\) and \(y=2\) simultaneously.

Answer

1. P\((\pi,-4)\)

2. Q\(\left(\frac{\pi}{3},2\right)\), R\(\left(\frac{5\pi}{3},2\right)\)

For more practice, try: Zeta Higher Textbook, Page 124, Exercise 8.6, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

For a \(2x\) trig equation from 0 to 360 degrees, expect 4 solutions.

Answer

\(x=30,150,210,330\)

For more practice, try: Zeta Higher Textbook, Page 121, Exercise 8.4, Questions 1(a), 1(b) and 2(a)

Solution

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Hint

For a \(2x\) trig equation from 0 to 360 degrees, expect 4 solutions.

Answer

\(x=120,150,300,330\)

For more practice, try: Zeta Higher Textbook, Page 121, Exercise 8.4, Questions 1(g), 2(b) and 2(g)

Solution

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Hint

Remember that \(\frac{\pi}{6}\) radians is equal to \(30\degree\).

Find \(x\degree-30\degree\) solutions then add \(30\degree\) to solve for \(x\).

Answer

\(x=\frac{11\pi}{12},\frac{23\pi}{12}\)

For more practice, try: Zeta Higher Textbook, Page 123, Exercise 8.5A, Questions 2(a), 2(c) and 2(d)

Solution

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Hint

Expect 4 solutions... Use a combination of skills from the previous questions.

Answer

\(x=\frac{\pi}{4},\frac{5\pi}{12}.\frac{5\pi}{4}.\frac{17\pi}{12}\)

No similar questions found in the Zeta textbook. Try this SQA exam question from 2021.

Solution

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7 Graph Transformations

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Hint

Make a list of all coordinates on the original graph (here there are 3).

Everything outside the \((x)\) bracket affects \(y\)-coordinates. These transformations are usually seen as intuitive.

Everything inside the \((x)\) bracket affects \(x\)-coordinates. These transformations are usually seen as counterintuitive.

Make sure your final graph has all the transformed images of the original coordinates clearly shown.

Answer

See full solution below for the graph and coordinates.

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Remember to apply the two transformations following the rules of operation: \(\times(-1)\) then \(-1\)

As well as changing the coordinates, think about how the transformations will change the shape and position of the graph. Here, \(-f(x)\) will flip the graph upside down.

Answer

See full solution below for the graph and coordinates.

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

If \(-f(x)\) is a vertical flip, what is \(f(-x)\)?

Answer

See full solution below for the graph and coordinates.

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Don't ignore the coordinate of \((0,0)\) on the original graph.

Answer

See full solution below for the graph and coordinates.

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

\(2-g(x)\) can be rewritten as \(-g(x)+2\).

Answer

See full solution below for the graph and coordinates.

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

\(\frac{1}{2}g(x)\) is a vertical compression by factor 2.

Answer

See full solution below for the graph and coordinates.

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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8 Vectors

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Hint

Find the components of vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{QR}\).

Make sure your collinearity statement is carefully and fully worded.

Draw a sketch to make (b) easier to see.

Answer

1. \(2\overrightarrow{PQ}=\overrightarrow{QR}\), valid statement (see solution).

2. \(1:2\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution (a)

Solution (b)

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Hint

Check which method your teacher taught you to use. One option is the section formula.

Answer

E\((5,-1,4)\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

1. Use \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\)

2. The formula sheet reminds you how to do this.

3. Use the formula \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos{\theta}\), rearranged and in terms of \(\overrightarrow{EM}\) and \(\overrightarrow{ER}\).

Answer

1. \(\overrightarrow{EM}\)=

   \[\begin{pmatrix} \phantom{-}2\\ -6\\ -1 \end{pmatrix}\]

   and \(\overrightarrow{ER}=\)

   \[\begin{pmatrix} -3\\ -2\\ \phantom{-}3 \end{pmatrix}\]

2. 3

3. \(84.3^{\circ}\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution (a)

Solution (b)

Solution (c)

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Hint

If a vector has a magnitude of \(k\), multiplying that vector by \(\frac{1}{k}\) will create a parallel vector with a length of \(1\) (a unit vector).

Answer

\(\begin{pmatrix} \frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3} \end{pmatrix}\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

1. Start by expressing vector BF as BA\(+\)AC\(+\)CF.

2. Start similar to the above hint.

3. The following approaches will help:

• For all vectors, \(\mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2\)

• If \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular then \(\mathbf{a}\cdot\mathbf{b}=0\)

• If the angle \(\theta\) between two vectors and their magnitudes are known, then \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos{\theta}\) can be used.

Answer

1. \(-\mathbf{u}+\mathbf{v}+\mathbf{w}\)

2. \(-\frac{2}{3}\mathbf{w}-\mathbf{v}+\mathbf{w}\)

3. 1. \(\frac{9}{2}\) (ii) \(\frac{9}{2}\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution (a)

Solution (b)

Solution (c)

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9 Differentiation II

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Hint

A function is increasing when its derivative is positive.

Answer

\(x>-2\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

A function is decreasing when its derivative is negative.

Remember to draw a sketch to solve a quadratic inequation!

Answer

\(-2<x<4\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Stationary points occur when the derivative is zero.

Answer

\(x=-5,x=3\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

The second derivative can be used to determine the nature of a stationary point.

Answer

Maximum turning point at \((-2,49)\)

Minimum turning point at \((3,-76)\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

A function that is strictly increasing across an interval will "begin" at its lowest value and increase towards its highest value as \(x\) increases.

Answer

Min value is \(-5\)

Max value is \(7\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Finding a value of \(x\) which minimises a function is very similar in process to finding a stationary point.

In its most convenient form, we might hope to find only one stationary point within the domain, and for that stationary point to be a minimum turning point.

Answer

The function is minimised when \(x=3\), with justification (see solution).

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

The hint to Q52 applies here, but this time we are looking to maximise the function.

Answer

Max value of \(R(x)\) when \(x=\frac{9}{2}\).

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

When sketching the derivative, stationary points on \(f(x)\) translate to \(roots\) on \(f'(x)\).

Remember that the derivative of a cubic (\(\dots x^3\dots\)) is a quadratic (\(\dots x^2\dots\)).

Answer

See the solution.

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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10 Polynomials

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Hint

Synthetic division may be used to calculate a remainder from a polynomial division.

Answer

\(7\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

The numbers along the bottom row of a synthetic division, to the left of the remainder, form the coefficients of the quotient. This is needed to fully factorise a polynomial.

Answer

\((x+1)(x+2)(x-4)\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

To solve a polynomial equation, aim to factorise.

Answer

\(x=2,x=-\frac{3}{2},x=-1\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

To find the \(x\)-intercepts (or roots) of a polynomial, set \(y=0\) and solve. Note that we can already see one solution on the graph, which may be used to begin a synthetic division.

Answer

A\((-2,0)\) and B\((-1,0)\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Synthetic division can be performed even if one or more coefficients is unknown. The remainder will contain \(k\), here... but we know what the remainder should be.

Answer

\(k=-5\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

The roots on the graph give the factors of the function.

Note that one of the roots is a repeated root.

The \(y\)-intercept can be used to determine the value of \(k\).

Answer

\(k=2,a=-1,b=3\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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11 Integration

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Hint

To integrate, increase the power by 1 and then divide by the new power.

Answer

\(\frac{x^4}{4}-2x^3+5x^2-7x+C\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Writing roots in index form is an important skill that should be practising until it becomes fluid.

Where factions lead to arithmetic that can't confidently be done without working... use the side of your page for working!

Answer

\(3x^4+4x^\frac{3}{2}+C\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

After integrating, substitution of \(2\) and \(1\) is required, then a subtraction.

Answer

\(\dfrac{15}{2}\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Integration using the equation of a curve can be used to find the area between it and the \(x\)-axis.

Answer

\(\dfrac{20}{3}\) square units

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

To find an area between curves, integrate using "\(\text{upper}-\text{lower}\)".

Answer

\(\dfrac{875}{12}\) square units

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Integration can be used to find the original function, \(f'(x)\). Since this integral should contain the constant of integration (\(+C\)), substituting suitable values will be needed to find the value of \(C\).

Answer

\(f(x)=x^3-4x^2+7x+4\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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12 Addition Formulae

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Hint

By finding the lengths of any missing sides using Pythagoras, trig values of \(p\) and \(q\) can be stated using SOHCAHTOA.

Expanding the addition formulae will then allow these values to be substituted.

Answer

\(\sin{p}=\dfrac{3}{5}\)

\(\sin{q}=\dfrac{3}{\sqrt{34}}\)

\(\sin{(p+q)}=\dfrac{27}{5\sqrt{34}}\)

\(\cos{(p+q)}=\dfrac{11}{5\sqrt{34}}\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution (a)

Solution (b)(i)

Solution (b)(ii)

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Hint

There are several ways to expand \(\cos{2r}\). Since both the values of \(\sin{r}\) and \(\cos{r}\) can be found using the right-angled triangle, you should be able to obtain the correct answer regardless of which you choose to use.

Answer

\(\cos{r}=\dfrac{5}{\sqrt{29}}\)

\(\cos{2r}=\dfrac{21}{29}\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Expand \(\sin{2x}\) using the formula provided.

Checking the equation is arranged equal to zero, look to factorise.

Answer

\(x=30,90,150,270\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

There are several ways to expand \(\cos{2x}\).

Since the equation also contains a \(\cos{x}\) term, choose the expandion containing only \(\cos{x}\).

Checking the equation is arranged equal to zero, look to factorise.

Answer

\(x=0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

There are several ways to expand \(\cos{2x}\).

Since the equation also contains a \(\sin{x}\) term, choose the expandion containing only \(\sin{x}\).

Checking the equation is arranged equal to zero, look to factorise.

Answer

\(x=0,\pi,\frac{7\pi}{6},\frac{11\pi}{6},2\pi\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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13 The Circle

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Hint

Check your formula sheet for each form of the equation of a circle.

Answer

Radius\(=\sqrt{20}\)

Centre\(=(6,-1)\)

Radius\(=2\)

Centre\(=(4,3)\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Once you know the centre and radius of a circle, use \((x-a)^2+(y-b)^2=r^2\) to state its equation. There is no need to expand the brackets.

Answer

Centre\(=(-5,2)\)

Radius\(=6\)

\((x+7)^2+(y-1)^2=9\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution

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Hint

Start by obtaining the radii and centres of any circle with equations given - this should be the first step for any circle question which requires an element of problem-solving.

Draw the radii of each circle on the sketch, and mark down the coordinates of each centre.

Answer

\((x+6)^2+(y-4)^2=16\)

\((-2,4)\)

For more practice, try: Zeta Higher Textbook, Page , Exercise , Questions

Solution